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        <p>​    花了好几个晚自习做完Datalab后，深感自己对CSAPP的第二章还不够深入与熟练，重新来对每个函数的实现进行分析，并参照书本进行考证。</p>
<a id="more"></a>

<h3 id="bitAnd"><a href="#bitAnd" class="headerlink" title="bitAnd"></a>bitAnd</h3><p>实现x&amp;y，只能使用~和|运算符，个人认为Datalab中很容易想的一道题目，很顺畅想到德摩根律来对与运算转化为或运算。</p>
<h3 id="getByte"><a href="#getByte" class="headerlink" title="getByte"></a>getByte</h3><p>要求我们从x中提取第n个字节，这个考察了基本的单位换算，1 byte=8 bit，要提取第几个字节，就要右移多少位，相当于右移$8<em>n$位，同时也考察了基本的位运算，$8</em>n$相当于$2^3*n$，即对n左移3位，然后将x右移与0xff相与即可。</p>
<h3 id="logicalShift"><a href="#logicalShift" class="headerlink" title="logicalShift"></a>logicalShift</h3><p>考察了逻辑移位与算术移位，基本上所有语言的移位操作都为算术移位，只需要记得有符号数若符号位为1，算术右移是补1，其他任何移位操作都是补0，同时考察了书中所讲补码非的位级表示(P66)，对于整形，-x与~x+1所得到的的结果是一样的。然后这里算术右移可能会导致前面几位都变成1，所以要找一个掩码让变成1的变回0，所以构造一个掩码，可以用1来移动到最高位，取反后就得到除最高位为0，其他都为1的数，然后你要让算术右移产生的1变成0，正好把这个数右移n-1位，就可以让1与0&amp;运算得到0，原来的数字与1&amp;运算不变，由于不能用-符号，所以只能右移n位之后，再左移一位，这里注意左移导致最后一位变成0，所以还要或上0x01。</p>
<h3 id="Bitcount"><a href="#Bitcount" class="headerlink" title="Bitcount"></a>Bitcount</h3><p>非常难想的一道题目，思想是分治法的逆用，由于不能直接统计1的个数，可以看看Hamming Weight的实现，首先分为2位1组，然后数出有多少个1之后再分为4位一组，将两位一组的1相加之后·····以此类推。所以一定要构造掩码。</p>
<h3 id="Bang"><a href="#Bang" class="headerlink" title="Bang"></a>Bang</h3><p>返回!x，注意这里!x运算得到的只有0和1，不为0x00的时候就为1，所以就是判断一个数的位级表达是不是都是0，考察了相反数，无论正数负数，它的相反数与自己本身进行或运算，符号位肯定为1，只有0x00不是，所以求补码的相反数之后与本身进行或运算再进行移位，比较符号位即可。</p>
<h3 id="tmin"><a href="#tmin" class="headerlink" title="tmin"></a>tmin</h3><p>送分题，返回最小的二进制整形补码，一定是0x80000000，构造即可。</p>
<p>这里也有一个考的很多的点，Tmin=-Tmax-1，书上有详细介绍。</p>
<h3 id="fitsBits"><a href="#fitsBits" class="headerlink" title="fitsBits"></a>fitsBits</h3><p>想知道x是否能被n位的二进制补码表示，这里考察了数位扩展，首先x肯定能被32位表示。数位扩展又被分为0扩展和符号扩展，我们可以思考就把x当做一个n位的数，进行符号扩展之后看看和原来的x有什么不同，然后考察了异或运算，异或可以判断两个数是否相等。</p>
<h3 id="divpwr2"><a href="#divpwr2" class="headerlink" title="divpwr2"></a>divpwr2</h3><p>考察乘除运算，这里是考察除以2的幂，书本p71，除以2的幂即移位运算，而且使用的是右移，无符号数用逻辑移位，有符号数用算术移位，而且除法运算总是舍入到0，所以对于无符号数直接移位即可，但是补码必须进行一系列操作，这里我们要的是除以2的幂，向上舍入。</p>
<p>操作：(x+(1&lt;&lt;k)-1)&gt;&gt;k</p>
<p>证明：</p>
<p><img src="https://i.loli.net/2021/04/15/TIZOyMS3exBoFUm.png" alt="BUp"></p>
<h3 id="negate"><a href="#negate" class="headerlink" title="negate"></a>negate</h3><p>求补码相反数，以上。</p>
<h3 id="isPositive"><a href="#isPositive" class="headerlink" title="isPositive"></a>isPositive</h3><p>判断符号位即可。注意0。</p>
<h3 id="isLessOrEqual"><a href="#isLessOrEqual" class="headerlink" title="isLessOrEqual"></a>isLessOrEqual</h3><p>看似是比较，实际考察补码减法，根据符号位判断即可。</p>
<h3 id="ilog2"><a href="#ilog2" class="headerlink" title="ilog2"></a>ilog2</h3><p>和bitcount类似的思路，把二进制数分为几段来求1的位置。</p>
<h3 id="float-eng"><a href="#float-eng" class="headerlink" title="float_eng"></a>float_eng</h3><p>考察对浮点数存储与运算，首先明确以32位为例，浮点数由1sign+8exp+23frac组成，其次NaN是exp都为1，frac非0的数段。</p>
<h3 id="float-i2f"><a href="#float-i2f" class="headerlink" title="float_i2f"></a>float_i2f</h3><p>细节最多也是最难的一题，几乎考察了对浮点数所有的知识点。</p>
<p>首先明确：</p>
<p>规格化数，E=exp-bias，bias对于单精度浮点数是127(为什么是127，大概是因为$2^7$-1)</p>
<p>非规格化数，exp=0x00，E=1-bias。</p>
<p>这里引用：</p>
<blockquote>
<p>首先把特殊情况0x0和0x80000000挑出来，因为移位是解决不了这两个的exp和frac。为什么0x80000000时，exp为 158 ？因为补码表示范围最大权重是0x4000000，E最大取到31，而根据公式exp = E + Bias(127)，所以此时exp = 31 + 127 = 158。<br>去掉两个特殊情况，当x为负数时，将其变成正数，便于后面的计算。<br>令计数器为30，因为符号位不看，阶码域不为全1，则最高位不为1。依次减少右移位数，直到移动到第一个 1 处(MSB)停下，此时 i 计为 E 的权重，可根据公式计算exp。<br>将第一个 1 前的 0 左移去掉，因为31-23=8，int 转换为 float 要损失 8 位的精度，所以先把x 右移8位按位与 1 得到 frac，再来看最后 8 位是否要进位。<br>原 x 进行掩码运算，只保留最后 8 位的有效数字，若超出 128 则进位，或者最高位是1，但是整体是奇数时也要向偶数进位，所以判断尾数是不是 1 。若进位则 frac + 1，如果frac 也突破 23 的限制了，那就给 E 加上 1 的权重，再把 frac 掩码运算一次 。<br>最后依次填空符号位阶码域小数域即可。</p>
</blockquote>

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